## 2D Motion due to Gravity

So far, we have only considered the motion of an object under gravity in one dimension (vertically).

###### \[\,\]Figure 3.2.1

Now, let us instead generalise even further and consider motion under gravity in two dimensions! For example, we could kick a ball at an arbitrary angle.

###### \[\,\]Figure 3.2.2

Let us in fact be even more general than that. Let us instead say that we *projectÂ *a **projectile** at an arbitrary angle. This therefore includes the possibility that we throw, kick, fire etc. any projectile (object) rather than specifically kicking a ball. The resulting path of the projectile is referred to as its **trajectory**.

If gravity was turned off, the velocity of the projectile would not change after being kicked, resulting in a straight line trajectory as the projectile travels into space.

With gravity turned on, however, the projectile instead “bends” back towards the Earth, resulting in the shape of the trajectory instead being what is known as a **parabola**.

###### \[\,\]Figure 3.2.3

What happens to our SUVAT vectors under such 2D motion? Well, let’s have a look!

###### \[\,\]Figure 3.2.4

By definition, the displacement vector points from the reference location to the projectile.

The velocity vector points along the particle’s trajectory.

The acceleration vector doesn’t change at all, as expected! Acceleration due to gravity always points downwards and always has a magnitude of \(9.8\mbox{ m/s}^2\) on Earth!

So, neither the magnitude nor the direction of the acceleration vector is changing, but what about the other two vectors? How do we calculate how much they have changed at any instance in time?

To do that, we need to decompose these vectors!

\begin{align}\vec{u} = \vec{u}_x + \vec{u}_y && \vec{v} = \vec{v}_x + \vec{v}_y \\ \vec{a} = \vec{a}_x + \vec{a}_y && \vec{s} = \vec{s}_x + \vec{s}_y\end{align}

Let’s see how each of these component behave.

###### \[\,\]Figure 3.2.5

Woah, thereâ€™s a lot going on there! Letâ€™s instead look at each type of vector individually so that we can more clearly see how each one changes over time.

First, the displacement vector \(\vec{s}\).

###### \[\,\]Figure 3.2.6

The \(x\) component, \(\vec{s_x}\), gradually increases in length with time. \(\vec{s}_y\), on the other hand, appears to behave just as \(\vec{s}\) would if we instead threw the projectile vertically upwards. Its length increases from zero to a maximum, at the projectile’s maximum height, before reducing to zero again when the projectile hits the ground.

Next, the velocity vector \(\vec{v}\). Do you notice anything interesting about one of the components?

###### \[\,\]Figure 3.2.7

That’s right, \(\vec{v}_x\) isn’t changing, as expected! The object is only accelerating in the \(y\) direction (due to gravity) and therefore we do not expect the velocity component in the \(x\) direction to change.

What about the \(y\) component? Again, it behaves just as \(v\) did when we looked at 1D motion. It gets gradually smaller as the object approaches the maximum height, at which point it becomes zero. Then, it gets gradually larger but in the opposite direction as the objects falls.

Finally, the acceleration vector \(\vec{a}\), and weâ€™ve left the easiest one for last!

###### \[\,\]Figure 3.2.8

As already mentioned, neither the magnitude nor the direction of \(\vec{a}\) changes as acceleration due to gravity is constant. Therefore, \(\vec{a}_x=0\) and \(\vec{a}_y=\vec{a}\)!

Now that we know what is happening visually, let us now figure out how to deal with our vector equations in 2D.

In 1D, we began by calculating things like \(\vec{a} \cdot \vec{i}\). What do we obtain for such quantities in 2D?Â Well, let’s check!

\begin{align}\vec{a}\cdot\vec{i} &=( a_x\vec{i} + a_y\vec{j})\cdot\vec{i}\\&=a_x(\vec{i}\cdot\vec{i}) + a_y(\vec{j}\cdot\vec{i}) \end{align}

Again, however, recall that

\[\vec{i}\cdot\vec{i} = \vec{j}\cdot\vec{j} = 1\]

\[\vec{i}\cdot\vec{j} = \vec{j}\cdot\vec{i} = 0\]

This therefore simplifies down to

\begin{align}\vec{a}\cdot\vec{i} &=a_x(1) + a_y(0)\\&=a_x \end{align}

This result is indeed true for all of our vectors.

\begin{align}\vec{u}\cdot\vec{i} =u_x && \vec{v}\cdot\vec{i} =v_x \\ \vec{a}\cdot\vec{i} =a_x && \vec{s}\cdot\vec{i} =s_x\end{align}

But let’s not forget, we are working in 2D! We can also work out what \(\vec{a}\cdot\vec{j}\) etc. is too! For exampleÂ

\begin{align}\vec{a}\cdot\vec{j} &=( a_x\vec{i} + a_y\vec{j})\cdot\vec{j}\\&=a_x(\overbrace{\vec{i}\cdot\vec{j}}^{=0}) + a_y(\overbrace{\vec{j}\cdot\vec{j}}^{=1})\\&= a_y \end{align}

Again, we would get the same for our other vectors

\begin{align}\vec{u}\cdot\vec{j} =u_y && \vec{v}\cdot\vec{j} =v_y \\ \vec{a}\cdot\vec{j} =a_y && \vec{s}\cdot\vec{j} =s_y\end{align}

When we were dealing with 1D motion, we then used such relations to replace our SUVAT vector equations with SUVAT scalar equations by dotting our equations with \(\vec{i}\). Again, let’s do that here too!

Recalling that the first SUVAT vector equation is given by

\[\vec{v} = \vec{u}+\vec{a}t\]

and dotting both sides with \(\vec{i}\), we getÂ

\begin{align}\vec{v}\cdot\vec{i} &= (\vec{u}+\vec{a}t)\cdot\vec{i}\\&=\vec{u}\cdot\vec{i}+(\vec{a}\cdot\vec{i})t\end{align}

Using the relations we just obtained, namely that \(\vec{v}\cdot\vec{i}=v_x\) etc., this simplifies to

\[v_x=u_x+a_xt\]

Again, however, we are working in 2D and we could have instead dotted both sides of this equation with \(\vec{j}\)! This instead would give

\begin{align}\vec{v}\cdot\vec{j} &= (\vec{u}+\vec{a}t)\cdot\vec{j}\\&=\vec{u}\cdot\vec{j}+(\vec{a}\cdot\vec{j})t\end{align}

Again, using our relations, this becomes

\[v_y=u_y+a_yt\]

It looks like therefore that, in 2D, each of our SUVAT vector equations will instead be replaced by *two*Â scalar equations!

Indeed, if we follow the same procedure for each equation (and again do the dot product twice for the last equation), we obtain the following

### Key Point

The *2D SUVAT scalar equations* are given by

\begin{align}\begin{gathered}v_x=u_x+a_x t \\ s_x=\left(\frac{u_x+v_x}{2}\right)t\\ s_x=u_x t+\frac{1}{2}a_x t^2 \\ v_x^2=u_x^2+2a_xs_x \end{gathered} && \begin{gathered}v_y=u_y+a_y t \\ s_y=\left(\frac{u_y+v_y}{2}\right)t\\ s_y=u_y t+\frac{1}{2}a_y t^2 \\ v_y^2=u_y^2+2a_ys_y \end{gathered} \end{align}

where \(\vec{s}=s_x\vec{i}+s_y\vec{j}\) etc.

In two dimensions, there are therefore in fact eight SUVAT equations, a set of four for the \(x\)-component vectors and a set of four for the \(y\)-component vectors!

Hence, if we are given any vector instead in *polar *form in a question on 2D projectile motion (or if the magnitude and angle are stated with words), we must first convert that vector from polar form into component form (using trigonometry as before) so that we may use them in these eight equations!

Of course, we can also further simplify these equations down in the case of projectile motion due to gravity. As we discussed, the acceleration vector always has a magnitude of \(g=9.8\mbox{ m/s}^2\) and always point in the negative \(y\) direction (using standard \(x\)-\(y\) axes). Therefore

\begin{align}a_x=0 && a_y=-g\end{align}

Note also that, upon inserting that \(a_x=0\) into the four \(x\) component SUVAT equations, they reduce to only two equations.

For example, both of the equations

\begin{align}v_x=u_x+a_xt\end{align}

\begin{align}v_x=u_x^2 + 2a_xs_x\end{align}

reduce to \(v_x=u_x\), i.e. that the \(x\) component of velocity is constant.

Likewise, the following two equations

\begin{align}s_x=\left(\frac{u_x+v_x}{2}\right)t\end{align}

\begin{align}s_x=u_xt+\frac{1}{2}a_xt^2\end{align}

reduce to \(s_x=u_xt\), which is a rearranged version of an equation that is only true when there is no acceleration, namely

\begin{align}\mbox{velocity} = \frac{\mbox{displacement}}{\mbox{time}}\end{align}

### Key Point

The 2D SUVAT scalar equations for projectile motion are given by

\begin{align}\begin{gathered}v_x=u_x\\s_x=u_x t \end{gathered} && \begin{gathered}v_y=u_y-g t \\ s_y=\left(\frac{u_y+v_y}{2}\right)t\\ s_y=u_y t-\frac{1}{2}g t^2 \\ v_y^2=u_y^2-2gs_y \end{gathered} \end{align}

where \(\vec{s}=s_x\vec{i}+s_y\vec{j}\) etc.

(Again, there is no need to memorise this. Simply insert that \(a_x=0\) and \(a_y=-g\) into the more general 2D SUVAT scalar equations.)

While having more equations to deal with may look intimidating at first, we use the exact same approach as we did in the 1D case. If we know three SUVAT quantities for either of the two sets of equations, we can then find a fourth (and then the fifth if necessary).

## Guided Example 3.2.1

A particle is projected from \(o\) on the horizontal ground with an initial velocity of \(3\vec{i}+5\vec{j}\mbox{ m/s}\).

After \(0.5\) seconds:

**(a)** What is the object’s velocity?

**(b)** What is the object’s acceleration?

**(c)** What is the object’s displacement from \(o\)?

In this question, we are given the initial velocity in component form, which is to our benefit! We don’t have to go through the hassle of decomposing the vector.

Instead, we immediately know what \(u_x\) and \(u_y\) is.

\begin{align} u_x= 3&& u_y=5\end{align}

**(a)** We are being asked to find \(\vec{v} = v_x\vec{i}+v_y\vec{j}\).

and therefore we must find both \(v_x\) and \(v_y\).

First, let us find \(v_x\). We should therefore use an \(x\) component SUVAT equation.

**Step 1: State the quantities that are given in the question and the quantities that we are looking for. Be careful with signs!**

We are given the following

\begin{align} u_x = 3 && a_x=0 && t=0.5 \end{align}

and we are looking for \(v_x\).

\[\,\]

**Step 2: Choose the most suitable SUVAT equation(s) based on Step 1.**

As we are not told \(s_x\) nor do we want to find \(s_x\), we should choose the \(x\) component SUVAT equation which does not have \(s_x\) in it, namely

\[v_x=u_x+a_xt\]

\[\,\]

**Step 3: Insert the quantities into the equation(s).**

Subbing in our known values, we obtain

\begin{align}v_x&=3+(0)(0.5)\\&=3 \mbox{ m/s}\end{align}

Next, let us find \(v_y\). We should therefore now use a \(y\) component SUVAT equation.

**Step 1: State the quantities that are given in the question and the quantities that we are looking for. Be careful with signs!**

We are given the following

\begin{align} u_y = 5 && a_y=-9.8 && t=0.5 \end{align}

and we are looking for \(v_y\).

\[\,\]

**Step 2: Choose the most suitable SUVAT equation(s) based on Step 1.**

As we are not told \(s_y\) nor do we want to find \(s_y\), we should choose the \(y\) component SUVAT equation which does not have \(s_y\) in it, namely

\[v_y=u_y+a_yt\]

\[\,\]

**Step 3: Insert the quantities into the equation(s).**

Subbing in our known values, we obtain

\begin{align}v_y&=5+(-9.8)(0.5)\\&=0.1 \mbox{ m/s}\end{align}

Therefore, the object’s velocity after \(0.5\) seconds is \(\vec{v} = 3\vec{i}+0.1\vec{j}\mbox{ m/s}\).

**(b)** As the object is accelerating due to gravity, the object’s acceleration will always be \(\vec{a} = -9.8 \vec{j}\mbox{ m/s}^2\).

**(c)** We are being asked to find \(\vec{s} = s_x\vec{i}+s_y\vec{j}\).

and therefore we must find both \(s_x\) and \(s_y\). First, let us find \(s_x\). We should therefore use an \(x\) component SUVAT equation.

**Step 1: State the quantities that are given in the question and the quantities that we are looking for. Be careful with signs!**

We are given the following

\begin{align} u_x = 3 && a_x=0 && t=0.5 \end{align}

and we are looking for \(s_x\).

\[\,\]

**Step 2: Choose the most suitable SUVAT equation(s) based on Step 1.**

As we are not told \(v_x\) nor do we want to find \(v_x\), we should choose the \(x\) component SUVAT equation which does not have \(v_x\) in it, namely

\[s_x=u_xt+\frac{1}{2}a_xt^2\]

\[\,\]

**Step 3: Insert the quantities into the equation(s).**

Subbing in our known values, we obtain

\begin{align}s_x &= 3(0.5)+\frac{1}{2}(0)(0.5^2)\\&= 1.5 \mbox{ m}\end{align}

Next, let us find \(s_y\). We should therefore now use a \(y\) component SUVAT equation.

We are given the following

\begin{align} u_y = 5 && a_y=-9.8 && t=0.5 \end{align}

and we are looking for \(s_y\).

\[\,\]

**Step 2: Choose the most suitable SUVAT equation(s) based on Step 1.**

As we are not told \(v_y\) nor do we want to find \(v_y\), we should choose the \(y\) component SUVAT equation which does not have \(v_y\) in it, namely

\[s_y=u_yt+\frac{1}{2}a_yt^2\]

\[\,\]

**Step 3: Insert the quantities into the equation(s).**

Subbing in our known values, we obtain

\begin{align}s_y &= 5(0.5)+\frac{1}{2}(-9.8)(0.5^2)\\&= 1.275 \mbox{ m}\end{align}

Therefore, the object’s displacement from \(o\) after \(0.5\) seconds is \(\vec{s} = 1.5\vec{i}+1.275\vec{j}\mbox{ m}\).

In some questions, we may be asked to find an \(x\) component quantity, e.g. \(s_x\), without seeming to have enough known \(x\) component quantities to use an \(x\) component equation. However, the question instead gives us many \(y\) component quantities. How can we use that to our advantage?

Well, are these two sets of equations somehow related? Yes – they both contain the time \(t\)!

As time is a scalar, it does not have an \(x\) and \(y\) component and therefore it is the same \(t\) that appears in all eight equations.

###### \[\,\]Figure 3.2.9

If we therefore know many \(y\) component quantities, we can first find the time \(t\) using these quantities. This then gives us an additional quantity to use in our \(x\) component equations!

## Guided Example 3.2.2

An object is launched with an initial velocity \(\vec{u} = x\vec{i} + 15 \vec{j}\mbox{ m/s}\).

If the object passes the point \(2\vec{i} + 3\vec{j}\mbox{ m}\), what are the two possible values of \(x\), correct to two decimal places?

In this question, we are given the following

\begin{align}u_y=15 && a_y=-9.8 && s_y=3\end{align}

and

\begin{align}a_x=0 && s_x=2\end{align}

and we are being asked to find \(u_x\).

As each SUVAT equation contains four variables, we need to know at least three \(x\) component quantities if we wish to use an \(x\) component equation. Currently, we only know two.

However, we do have enough \(y\) component quantities to use a \(y\) component equation to find \(t\). This will then become our third known \(x\) component quantity which will then allow us to find \(u_x\)!

Therefore, let us start by using a \(y\) component equation to find \(t\).

We are given the following

\begin{align} u_y = 15 && a_y=-9.8 && s_y=3 \end{align}

and we are looking for \(t\).

\[\,\]

**Step 2: Choose the most suitable SUVAT equation(s) based on Step 1.**

As we are not told \(v_y\) nor do we want to find \(v_y\), we should choose the \(y\) component SUVAT equation which does not have \(v_y\) in it, namely

\[s_y = u_yt+\frac{1}{2}a_yt^2\]

\[\,\]

**Step 3: Insert the quantities into the equation(s).**

Subbing in our known values, we obtain

\begin{align}3=15t+\frac{1}{2}(-9.8)t^2\end{align}

\begin{align}\downarrow\end{align}

\begin{align}4.9t^2-15t+3=0\end{align}

This can be solved using the quadratic formula to give

\begin{align}t&=\frac{15\pm\sqrt{(-15)^2-4(4.9)(3)}}{2(4.9)}\\&\approx 1.53 \pm 1.32\end{align}

Therefore, our two possible value for \(t\) are \(0.21\) and \(2.85\) seconds.

Now that we know our two values for \(t\), let us continue by using these values in an \(x\) component equation to find \(u_x\).

We are given the following

\begin{align} a_x=0 && s_x=2 && t=0.21 \mbox{ or } 2.85\end{align}

and we are looking for \(u_x\).

\[\,\]

**Step 2: Choose the most suitable SUVAT equation(s) based on Step 1.**

As we are not told \(v_x\) nor do we want to find \(v_x\), we should choose the \(x\) component SUVAT equation which does not have \(v_x\) in it, namely

\[s_x = u_xt+\frac{1}{2}a_xt^2\]

\[\,\]

**Step 3: Insert the quantities into the equation(s).**

First, let us start with \(t=0.21\) s. Subbing in our known values, we obtain

\begin{align}2=u_x (0.21)+\frac{1}{2}(0)(0.21^2)\end{align}

\begin{align}\downarrow\end{align}

\begin{align}u_x&=\frac{2}{0.21}\\&\approx 9.52 \mbox{ m/s}\end{align}

Next, let us instead use \(t=2.85\) s. Subbing in our known values, we obtain

\begin{align}2=u_x (2.85)+\frac{1}{2}(0)(2.85^2)\end{align}

\begin{align}\downarrow\end{align}

\begin{align}u_x&=\frac{2}{2.85}\\&\approx 0.70 \mbox{ m/s}\end{align}

Therefore, the two possible values of \(x\) as defined in the question are \(9.52\mbox{ m/s}\) and \(0.70\mbox{ m/s}\).

(The reason why there are two possible values of \(u_x\) is because trajectories intersect. This will be explained in more detail in the next lesson!)

As with 1D motion, we can also have questions where we are given information in “chunks” of time and which thereby involves the use of connected variables.

In the 1D case, we considered both the first chunk as well as the total journey so that the initial velocity component \(u=U\) was the same in both cases. This then lead to two simultaneous equation which would could then solve.

In the 2D case, we follows the exact same procedure. We consider both the first chunk as well as the total trajectory so that the initial velocity components \(u_x=U_x\) and \(u_y=U_y\) are the same in both cases. This therefore means that we will now instead obtain *four* simultaneous equations which we can then solve.

## Guided Example 3.2.3

An object is projected from \(o\) on the horizontal ground. After \(t_1\) seconds, it is displaced \(2\vec{i} + 5\vec{j}\) metres from \(o\). After another \(t_2\) seconds, it is displaced \(3\vec{i} + 4\vec{j}\) metres from \(o\).

**(a)** Find the values of \(t_1\) and \(t_2\), correct to two decimal places.

**(b)** Find the object’s initial velocity, correct to two decimal places.

**(a)** This is a question involving uniform acceleration in “chunks”.

We should therefore look at the first \(t_1\) seconds as well as the total \(t_1+t_2\) seconds, so that \(u_x=U_y\) and \(u_y = U_y\) in both cases.

In each case, we will therefore write down both an \(x\) component and a \(y\) component equation, thereby resulting in a total of four equations.

Let’s start with \(x\) components for the first \(t_1\) seconds.

We are given the following

\begin{align} s_x = 2 && a_x = 0\end{align}

and we are looking for \(u_x=U_x\) (as it’s a connected variable) and \(t=t_1\) (as it’s what we’re looking for).

\[\,\]

**Step 2: Choose the most suitable SUVAT equation(s) based on Step 1.**

As we are not told \(v_x\) nor do we want to find \(v_x\), we should choose the SUVAT equation which does not have \(v_x\) in it, namely

\[s_x=u_xt+\frac{1}{2}a_xt^2\]

\[\,\]

**Step 3: Insert the quantities into the equation(s).**

Subbing in our known values, we obtain

\begin{align}2=U_x t_1+\frac{1}{2}(0)t_1^2\end{align}

\begin{align}\downarrow\end{align}

\begin{align}2=U_x t_1\end{align}

Next, let’s look at the \(y\) components for the first \(t_1\) seconds.

We are given the following

\begin{align} s_y = 5 && a_y = -9.8\end{align}

and we are looking for \(u_y=U_y\) (as it’s a connected variable) and \(t=t_1\) (as it’s what we’re looking for).

\[\,\]

**Step 2: Choose the most suitable SUVAT equation(s) based on Step 1.**

As we are not told \(v_y\) nor do we want to find \(v_y\), we should choose the SUVAT equation which does not have \(v_y\) in it, namely

\[s_y=u_yt+\frac{1}{2}a_yt^2\]

\[\,\]

**Step 3: Insert the quantities into the equation(s).**

Subbing in our known values, we obtain

\begin{align}5=U_y t_1+\frac{1}{2}(-9.8)t_1^2\end{align}

\begin{align}\downarrow\end{align}

\begin{align}5 = U_y t_1 -4.9t_1^2\end{align}

Next, let’s look at the \(x\) components after \(t_1+t_2\) seconds.

We are given the following

\begin{align} s_x = 3 && a_x = 0\end{align}

and we are looking for \(u_x=U_x\) (as it’s a connected variable) and \(t=t_1+t_2\) (as it’s what we’re looking for).

\[\,\]

**Step 2: Choose the most suitable SUVAT equation(s) based on Step 1.**

As we are not told \(v_x\) nor do we want to find \(v_x\), we should choose the SUVAT equation which does not have \(v_x\) in it, namely

\[s_x=u_xt+\frac{1}{2}a_xt^2\]

\[\,\]

**Step 3: Insert the quantities into the equation(s).**

Subbing in our known values, we obtain

\begin{align}3=U_x (t_1+t_2)+\frac{1}{2}(0)(t_1+t_2)^2\end{align}

\begin{align}\downarrow\end{align}

\begin{align}3 =Â U_x (t_1+t_2)\end{align}

Finally, let’s look at the \(y\) components after \(t_1+t_2\) seconds.

We are given the following

\begin{align} s_y = 4 && a_y = -9.8\end{align}

and we are looking for \(u_y=U_y\) (as it’s a connected variable) and \(t=t_1+t_2\) (as it’s what we’re looking for).

\[\,\]

**Step 2: Choose the most suitable SUVAT equation(s) based on Step 1.**

As we are not told \(v_y\) nor do we want to find \(v_y\), we should choose the SUVAT equation which does not have \(v_y\) in it, namely

\[s_y=u_yt+\frac{1}{2}a_yt^2\]

\[\,\]

**Step 3: Insert the quantities into the equation(s).**

Subbing in our known values, we obtain

\begin{align}4=U_y (t_1+t_2)+\frac{1}{2}(-9.8)(t_1+t_2)^2\end{align}

\begin{align}\downarrow\end{align}

\begin{align}4 =Â U_y (t_1+t_2) -4.9(t_1+t_2)^2\end{align}

We therefore have the following four equations

\begin{align}2=U_x t_1\end{align}

\begin{align}5 = U_y t_1 -4.9t_1^2\end{align}

\begin{align}3 = U_x(t_1+t_2)\end{align}

\begin{align}4 = U_y (t_1+t_2) -4.9(t_1+t_2)^2\end{align}

We can reduce this to two equations by inserting into the second equation that \(t_1=\displaystyle\frac{2}{U_x}\) from the first equation, and inserting into the fourth equation that \(t_1+t_2=\displaystyle\frac{3}{U_x}\) from the third equation, leaving us with

\begin{align}5 = \frac{2 U_y}{U_x} -\frac{19.6}{U_x^2}\end{align}

\begin{align}4 = \frac{3U_y}{U_x}Â -\frac{44.1}{U_x^2}\end{align}

We can remove the “\(\displaystyle\frac{U_y}{U_x}\)” terms by multiplying the first equation by \(3\) and the second equation by \(2\)

\begin{align}15 = \frac{6 U_y}{U_x} -\frac{58.8}{U_x^2}\end{align}

\begin{align}8 = \frac{6U_y}{U_x}Â -\frac{88.2}{U_x^2}\end{align}

and then subtracting one equation from the other

\begin{align}7 = \frac{29.4}{U_x^2}\end{align}

Solving for \(U_x\), we obtain

\begin{align}U_x &= \sqrt{\frac{29.4}{7}}\\&\approx 2.05 \mbox{ m/s}\end{align}

We can now easily find \(t_1\) by rearranging the following

\begin{align}2=U_x t_1\end{align}

to obtain

\begin{align}t_1 &= \frac{2}{U_x}\\&=\frac{2}{2.05}\\&\approx 0.98 \mbox{ s}\end{align}

Now, we can easily find \(U_y\) by rearranging

\begin{align}5 = U_y t_1 -4.9t_1^2\end{align}

to obtain

\begin{align}U_y&=\frac{5+4.9t_1^2}{t_1}\\&=\frac{5+4.9(0.98^2)}{0.98}\\&\approx 9.90 \mbox{ m/s}\end{align}

Finally, we can find \(t_2\) by using e.g.Â

\begin{align}3 = U_x(t_1+t_2)\end{align}

to obtain

\begin{align}t_2 &= \frac{3-U_x t_1}{U_x}\\&=\frac{3-(2.05)(0.98)}{2.05}\\&\approx 0.48 \mbox { s}\end{align}

Hence the values of \(t_1\) and \(t_2\) are \(0.98\) seconds and \(0.48\) seconds respectively.

**(b)Â **As we found that \(U_x=2.05\mbox{ m/s}\) and \(U_y=9.90\mbox{ m/s}\), the initial velocity is therefore

\begin{align}\vec{u} = 2.05\vec{i} + 9.90\vec{j} \mbox{ m/s}\end{align}

Also, as was the case with 1D motion, we can also have questions involving multiple *objects*. For such questions involving projectile motion, it is most common for one of the two objects to be stationary, thereby being referred to as a *target*. We shall look at examples of such a scenario in the next lesson.

Finally, with 1D motion, we also considered the concept of *partial variables *to deal with an instantaneous acceleration change. However, with projectile motion, this concept does not appear necessary as acceleration due to gravity never changes and is instead always \(\vec{a}=-9.8 \vec{j}\mbox{ m/s}^2\)!

However, in a later topic, we will allow projectiles to *bounce* off the ground which does indeed involve an instantaneous acceleration change. More on that later!